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If $x \geq \min\{y,w\}$ then $x \leq y+w$.

This is kind intuitive,very trivial. Or not? It's like triangular inequality, isn't it?

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$x \leq y+x$ is just $0 \leq y$, no? – J. M. Aug 14 '12 at 13:31
Now I´s $x\leq y+w$.My bad. – HipsterMathematician Aug 14 '12 at 13:35
@MeAndMath It is still false as stated. See sebigu's example below. – Derek Allums Aug 14 '12 at 13:37
Every $x$ sufficiently big will disprove your conjecture. – enzotib Aug 14 '12 at 13:46
up vote 3 down vote accepted

This is wrong. Take $y,w=1$ and $x=3$.

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Yes,yes,yes.If you considerate $y,w$ to be equal it´s wrong.Right.I forgot that possibility...what if $y\neq w$ ? – HipsterMathematician Aug 14 '12 at 13:38
Then take $y=0, w=1$ and $x=2$. – Derek Allums Aug 14 '12 at 13:39
Take $y=2,w=3$ and $x=6$. – sebigu Aug 14 '12 at 13:40
GREAT!THANKS IN ADVANCE! – HipsterMathematician Aug 14 '12 at 13:40
No problem. Please mark this questing as solved. – sebigu Aug 14 '12 at 13:42

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